Finding Angle of Sphere Falling From Table Edge

[voko]

But $\omega=\frac { v }{ r }$

Is that true when the ball is slipping?

 

[Satvik Pandey]

Yes, you can apply the integrating factor method at the end.
Explain your method first, and define the variables. What are v, a, and alpha? What is omega? What do you mean that you can not use a=rα? And do not use the symbol a for two different things (acceleration of the CoM and v².

Let $\theta$ be the angle at which the ball is slipping and rotating together.
The CoM of the sphere performs circular motion about the edge. right?
Let the tangential velocity and the acceleration of the CoM at that particular angle $\theta$ be $v$ and $a$ respectively.
Let $\omega$ be the angular velocity of the CoM of the ball about the edge.

What do you mean that you can not use a=rα? And do not use the symbol a for two different things (acceleration of the CoM and v².

I meant $\alpha$ to be the angular acceleration of the that ball about the CoM. We can not use this formula as the ball is slipping.

Sorry for using a for two different things. I am going to change it.

Is that true when the ball is slipping?

Here v is the tangential velocity of the CoM at angle $\theta$ and $\omega$ is the angular velocity of the CoM of the ball and the ball is doing circular motion about the edge. So can we not use this equation $v=r \omega$?

 

[Satvik Pandey]

So I have 2 equations-

$mgcos\theta -N=m\frac { { v }^{ 2 } }{ r }$

and $\\ mgsin\theta -\mu N=ma$

$mgcos\theta -m\frac { { v }^{ 2 } }{ r }= N$

or and $\\ mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r })=ma$

Here we can not use $a=r \alpha$

So $\\ mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r })=m\frac{dv}{dt}$

or $\\ mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r })=m\frac{dv}{d \theta} \omega$

But $\omega=\frac { v }{ r }$

So $mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r } )=m\frac { vdv }{ rd\theta }$

Let ${ v }^{ 2 }=x\\ So 2vdv=dx$

$gsin\theta -\mu (gcos\theta -\frac { x }{ r } )=\frac { dx }{ 2rd\theta }$

$\frac { dx }{ 2rd\theta } =gsin\theta -\mu gcos\theta +\mu \frac { x }{ r }$

$\frac { dx }{ d\theta } -2\mu x+2rg(\mu cos\theta -sin\theta )=0$

I think this is linear differential equation. Should I solve this with integrating factor method?
Am I correct till here?

I have now edited it. Please see the quoted text.

 

[voko]

Ah, so $\omega$ is the angular velocity of the CoM, not of the ball itself. Disregard #91 in that case.

 

[Satvik Pandey]

Ah, so $\omega$ is the angular velocity of the CoM, not of the ball itself. Disregard #91 in that case.

Is differential equation that I have derived correct?

 

[voko]

I do not see any obvious mistake.

 

[ehild]

Let $\theta$ be the angle at which the ball is slipping and rotating together.

Theta is the angle of the radius, connecting the centre of the sphere with the edge, with respect to the vertical.
As the sphere touches the edge, the centre moves along a circle, but that circular motion is decoupled from the rotation of the sphere.

The CoM of the sphere performs circular motion about the edge. right?
Let the tangential velocity and the acceleration of the CoM at that particular angle $\theta$ be $v$ and $a$ respectively.
Let $\omega$ be the angular velocity of the CoM of the ball about the edge.

that is correct.

I meant $\alpha$ to be the angular acceleration of the that ball about the CoM. We can not use this formula as the ball is slipping.

You can totally disregard the rotation of the ball about its centre because of the slipping. The slipping ensures that the force of friction is kinetic, wirth defined magnitude Fk=μN.
But you can define angular variables for the circular motion of the centre. The angular velocity of the circular motion is ω=v/r, and the angular acceleration is α=a/r.

 

[ehild]

I have now edited it. Please see the quoted text.

The editing looks terrible, :Dbut the equation is OK.

 

[Satvik Pandey]

The editing looks terrible, :Dbut the equation is OK.

It is very lengthy to solve that equation with Integrating Factor method.
I hope that I solve it soon.

 

[ehild]

There is an other method to solve the linear inhomogeneous differential equation x'-2μx=f(θ).
Take the general solution of the homogeneous part: it is x=Ce^2μθ.
Add a particular solution X_p of the inhomogeneous equation: the general solution will be x=Ce^2μθ+X_p.
In case the inhomogeneous part contains sine and cosine functions, you can seek the solution in the form Xp=Acos(θ)+Bsin(θ). Replacing X_p =Acos(θ)+Bsin(θ) into the equation, you can find the constants A and B.

 

[Satvik Pandey]

There is an other method to solve the linear inhomogeneous differential equation x'-2μx=f(θ).
Take the general solution of the homogeneous part: it is x=Ce^2μθ.
Add a particular solution X_p of the inhomogeneous equation: the general solution will be x=Ce^2μθ+X_p.
In case the inhomogeneous part contains sine and cosine functions, you can seek the solution in the form Xp=Acos(θ)+Bsin(θ). Replacing X_p =Acos(θ)+Bsin(θ) into the equation, you can find the constants A and B.

I am getting

$x=-rg\left( \frac { 3sin\theta -cos\theta }{ 2 } \right) +C$

or ${ v }^{ 2 }=-rg\left( \frac { 3sin\theta -cos\theta }{ 2 } \right) +C$

Am I correct?

 

[ehild]

I do not think so. Have you tried to substitute it back into the de?

 

[Satvik Pandey]

I do not think so. Have you tried to substitute it back into the de?

What is de?

 

[ehild]

differential equation

 

[Satvik Pandey]

differential equation

No

 

[ehild]

Try.
And read my post #100 carefully.

 

[Satvik Pandey]

There were some mistakes in my solution. Now I got
$x=-\frac { rg }{ 2 } (3sin\theta +cos\theta )+C$.
I have substituted $\mu=0.5$.

 

[ehild]

What is C? Is X solution if C is not zero?

 

[Satvik Pandey]

Should I find $x$ at angle 41.82 and put it in this equation to find C?

 

[ehild]

Yes, it would be the next step if your solution were correct. But it is not. Read my post ##100.

 

[Satvik Pandey]

I was trying to solve it using IF method.:p

$\frac { dx }{ d\theta } -x=2rg(sin\theta -\mu cos\theta )$

Here $IF=e^{\int-d\theta}=e^{-\theta}$

$x{ e }^{ -\theta }=\int { { e }^{ -\theta } } \left\{ 2rg(sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } \left\{ (sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } \left\{ (sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } sin\theta \quad d\theta -\mu \int { { e }^{ -\theta }cos\theta } d \theta$

$x{ e }^{ -\theta }=2rg\left\{ -\frac { { e }^{ -\theta }\left( sin\theta +cos\theta \right) }{ 2 } -\frac { 1 }{ 2 } \frac { { e }^{ -\theta }\left( sin\theta -cos\theta \right) }{ 2 } \right\}$

$x{ e }^{ -\theta }=-rg{ e }^{ -\theta }\left\{ \frac { \left( sin\theta +cos\theta \right) }{ 1 } +\frac { \left( sin\theta -cos\theta \right) }{ 2 } \right\}$

$x{ e }^{ -\theta }=-rg{ e }^{ -\theta }\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\}$

$x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +C$

What am doing wrong?

 

[ehild]

I was trying to solve it using IF method.:p

$\frac { dx }{ d\theta } -x=2rg(sin\theta -\mu cos\theta )$

Here $IF=e^{\int-d\theta}=e^{-\theta}$

$x{ e }^{ -\theta }=\int { { e }^{ -\theta } } \left\{ 2rg(sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } \left\{ (sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } \left\{ (sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } sin\theta \quad d\theta -\mu \int { { e }^{ -\theta }cos\theta } d \theta$

$x{ e }^{ -\theta }=2rg\left\{ -\frac { { e }^{ -\theta }\left( sin\theta +cos\theta \right) }{ 2 } -\frac { 1 }{ 2 } \frac { { e }^{ -\theta }\left( sin\theta -cos\theta \right) }{ 2 } \right\}$

Where is the integration constant? You must add it before multiplying by e^θ.
$x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +C$
is not the general solution of the differential equation. You can see if you substitute it back.

 

[Satvik Pandey]

Where is the integration constant? You must add it before multiplying by e^θ.
$x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +C$
is not the general solution of the differential equation. You can see if you substitute it back.

Oh! My silly mistake.
$x{ e }^{ -\theta }=-rg{ e }^{ -\theta }\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +C$

or $x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +\frac { C }{ { e }^{ -\theta } }$

or $x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +{ Ce }^{ \theta }$. Right?

 

[ehild]

It is right now.

Ce^θ is the general solution of the homogeneous equation x'-x=0. I said earlier that the general solution is equal to that + a particular solution of the original equation. You repeatedly ignored my advice to look at my post #100 or substitute back into the original equation.

 

[Satvik Pandey]

You repeatedly ignored my advice to look at my post #100 or substitute back into the original equation.

Sorry:s.
So I should now find x at angle 41.82 and put it in this equation to find C.
At $\theta=0$ velocity of the ball is also 0. Can I also use this condition to find C?

 

[ehild]

Sorry:s.
So I should now find x at angle 41.82 and put it in this equation to find C.

Yes. But take care, you should use the angle in radians in the exponent.

At $\theta=0$ velocity of the ball is also 0. Can I also use this condition to find C?

No, it was a different situation, with Fs not equal to μN except the endpoint. You get v at 41.82° from that solution.

 

[Satvik Pandey]

No, it was a different situation, with Fs not equal to μN except the endpoint.

Fs is static friction. right?
Do you want to say that till angle 41.82 static friction acts on the ball and after that kinetic friction acts on the the ball and in differential equation we have assumed $\theta$ to be the angle at which the ball is slipping and rotating together that's why we can't use the initial condition at which v and $\theta$ are equal to zero.right?
But the value of $\mu_{s}= \mu_{k}=0.5$ so numerically Frictional force is always equal to 0.5N.

 

[ehild]

Fs is static friction. right?
Do you want to say that till angle 41.82 static friction acts on the ball and after that kinetic friction acts on the the ball and in differential equation we have assumed $\theta$ to be the angle at which the ball is slipping and rotating together that's why we can't use the initial condition at which v and $\theta$ are equal to zero.right?

In the second stage of the motion, yes.

But the value of $\mu_{s}= \mu_{k}=0.5$ so numerically Frictional force is always equal to 0.5N.

No. The static friction is a limit, not a defined value. Fs≤μ_sN while the kinetic friction is Fk=μ_kN.When the ball rolled we used the condition that the angular velocity of rotation was the same as the angular velocity of the circular motion the CoM performed.

 

[voko]

$\frac { dx }{ d\theta } -x=2rg(sin\theta -\mu cos\theta )$

In #93, the equation was $$ \frac { dx }{ d\theta } -2 \mu x + 2rg(\mu cos\theta - sin\theta) = 0. $$

 

[voko]

One thing to keep in mind is that the general solution of the equation $$ {dx \over ds} - k x = f(s) $$ is given by $$ x = e^{ks} \left[x_0 + \int\limits_0^s e^{-kt} f(t) dt\right] .$$